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2 years ago

Plot the foci of the hyperbola represented by the equation x^2/625 − y^2/3600 =1 100pts

Mathematics

1 answer:

Pavel [41]2 years ago

3 0

The attached image represents the foci of the hyperbola

<h3>How to determine the foci?</h3>

The equation of the hyperbola is given as:

$\frac{x^2}{625} - \frac{y^2}{3600} = 1$

Rewrite as:

$\frac{x^2}{25^2} - \frac{y^2}{60^2} = 1$

A hyperbola is represented as:

$\frac{(x - h)^2}{b^2} - \frac{(y - k)^2}{a^2} = 1$

This means that:

h = 0

k = 0

b = 25

a = 60

Next, calculate c the distance from the center to the focus using:

$c = \sqrt{a^2 -b^2}$

This gives

$c = \sqrt{60^2 -25^2}$

Evaluate

$c = \pm \sqrt{2975}$

This means that:

Foci = (0, -√2975) and (0, √2975)

See attachment for the hyperbola and the foci

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

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Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$